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Suppose that $z=xy$ is in the centre where $x\in N$ and $y\in K$. Then for all $u\in K$, $uxy=xyu$. But $uxy=\phi(u)(x)uy$ so that $x=\phi(u)(x)$ (and $uy=yu$). As this is true for all $u\in K$ then b …

Theo's example works for question (2) as well. Let $p$ be a prime (it doesn't have to be $3$). Let $\alpha:Z_{p^2}\times Z_p\to Z_p\times Z_p$ be the projection map and $\beta=\alpha$. Then the fibre …

Let $G$ be the group of affine linear maps over the Galois field $k=GF(16)$ of order $16$. The elements of $G$ are maps from $k$ to itself of the form $x\mapsto ax+b$ where $a\in k^*$ and $b\in G$. Th …

The answer to your question is "there must be, it's just a question of doing the bookkeeping carefully". It's well-known that a subgroup of $\mathrm{PGL}(2,p)$ with order prime to $p$ is either cyclic …

A transitive subgroup $G$ of $S_p$ contains a Sylow $p$-subgroup $P$ having order $p$. If it has only the one, then $P$ is normal in $G$ and so $G$ lies in the normalizer $N$ of $P$ in $S_p$. This is …

For your first question, I presume you also wish to insist that $k$ be the least integer such that $A^k=I$. The matrix $A$ is then similar over your field to a direct sum $B_1,\ldots,B_m$ of companion …